Part II Problems and Solutions
Problem 1: [Complex numbers, roots of unity]
(a) Make a table with three columns. Each row will contain three representations of a
complex number z: the “rectangular” expression z = a + bi (with a and b real); the “polar”
expression |z|, Arg(z); and a little picture of the complex plane with the complex number
marked on it. There are five rows, containing, in one column or another, the following
complex numbers:
(i) 1 − i
(ii) z such that |z| = 2 and Arg(z) = π/6
(iii) A square root of i with negative real and imaginary parts
(iv) A sixth root
1 with argument θ such that 0 < θ < π /2
√ of
−
13
(v) ((1 + i )/ 2) .
(b) Find the complex roots of the following equations: z4 + 4 = 0; z2 + 2z + 2 = 0.
Solution: (a) The rectangular expression gives the coordinates for the little pictures. Any
angle may be altered by adding a multiple of 2π. The table below does not have a column
for the pictures – you will have to supply those. The last column is for problem 2(a)
√1 − i
3 + i√
(−1 −√i )/ 2
(1 + 3i )√
/2
(−1 + i )/ 2
(b) (i) ±1 ± i; or
√
√
2, −π /4
2, π/6
1, 5π /4
1, π/3
1, 3π/4
√
2e−πi/4
2eπi/6
e5πi/4
eπi/3
e3πi/4
2ekπi/4 where k = 1, 3, 5, 7. (ii) −1 ± i.
Problem 2: [Complex exponentials, Sinusoids]
(a) Add a fourth column to table you made in 1(a) by giving the exponential representation
z = Aeiθ (with A and θ real).
(b) Find all complex numbers z = a + bi such that ez = −2.
(c) Find an expression for cos(4t) in terms of sums of powers of sin t and cos t by using
(eit )4 = e4it and Euler’s formula.
(d) The Mathlet Complex Exponential will probably be useful in understanding the rest
of this problem. Open it and explore its functionalities. The Help button lists most of
them. Notice that in the left window, the real part a ranges between −1 and 1, while the
imaginary part b ranges from −8 to 8. You use the left-hand window to pick out a complex
Part II Problems and Solutions
OCW 18.03SC
number a + bi. When you do, a portion of the line through it and zero is drawn. This line
is parametrized by ( a + bi )t. At the same time, the curve parametrized by the complexvalued function e(a+bi)t is drawn on the right window.
For each of the following functions f (t), make a sketch of the graph (in a convenient range),
find a value of w = a + bi such that Re(ewt ) = f (t), and sketch the trajectory of the complex
valued function ewt (i.e. its set of values).
(i) f (t) = cos(2πt)
(ii) f (t) = e−t
(iii) f (t) = e−t cos(2πt).
(iv) The constant function with value 1.
Solution: (a) The solution in the right hand column of the table above.
(b) e a+bi = e a ebi so |e a+bi | = |e a ||ebi | = e a . Since | − 2| = 2, a = ln 2. Arg(e a+bi ) = b up to
adding multiples of 2π. Arg(−1) = π, so b is any odd multiple of π. Answer: ln 2 + bπi,
b = ±1, ±3, . . ..
(c) cos(4t) = Re(e4it ) = Re((eit )4 ) = Re((cos t + i sin t)4 ). By the binomial theorem, ( a +
bi )4 = a4 + 4a3 bi − 6a2 b2 − 4ab3 i + b4 , so we find cos(4t) = cos4 t − 6 cos2 t sin2 t + sin4 t.
(d) (i) w = 2πi. The trajectory is the unit circle.
(ii) w = −1. The trajectory is the positive real axis.
(iii) w = −1 + 2πi. The trajectory is a spiral, spiralling in towards the origin in a counter­
clockwise direction and passing though 1.
(iv) w = 0. The trajectory is the single point 1.
2
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18.03SC Differential Equations
Fall 2011
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